A random walk, in retrospect, looks like like directional movement at a speed of √n.

I aint clicking on LW links on my day off (ty for your service though). I am trying to reverse engineer wtf this poster is possibly saying though. My best guess: If we have a random walk in Z_2, with X_i being a random var with 2 outcomes, -1 or +1 (50% chance left at every step, 50% chance for a step to the right), then the expected squared distance from the origin after n steps E[ (Σ_{i=1}^n X_i)^2 ] = E[Σ_{i=1}^n X_i^2}] + E[Σ_{i not = j, i,j both in {1,2,…n}} X_i X_j}]. The expectation of any product E[X_i X_j] with i not = j is 0, (again 50% -1, 50% +1), so the 2nd expectation is 0, and (X_i)^2 is always 1, hence the whole expectation of the squared distance is equal to n => the expectation of the nonsquared distance should be on the order of root(n). (I confess this rather straightforward argument comes from the wikipedia page on simple random walks, though I swear I must have seen it before decades ago.)

Now of course, to actually get the expected 1-norm distance, we need to compute E[Σ_{i=1}^n |X_i| ]. More exciting discussion here if you are interested! https://mathworld.wolfram.com/RandomWalk1-Dimensional.html

But back to the original posters point… the whole point of this evaluation is that it is directionLESS, we are looking at expected distance from the origin without a preference for left or right. Like I kind of see what they are trying to say? If afterwards I ignored any intermediate steps of the walker and just looked at the final location (but why tho), I could say "the walker started at the origin and now is approx root(2n/pi) distance away in the minus direction, so only looking at the start and end of the walk I would say the average velocity is d(position)/(d(time)) = ( - root(2n/pi) - 0) /( n ) -> the walker had directional movement in the minus direction at a speed of root(2/(pi*n)) "

wait, so the “speed” would be O(1/root(n)), not root(n)… am I fucking crazy?